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Ex 6.3, 8 - Find a point on y = (x-2)2, tangent is parallel

Ex 6.3,8 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.3,8 - Chapter 6 Class 12 Application of Derivatives - Part 3
Ex 6.3,8 - Chapter 6 Class 12 Application of Derivatives - Part 4

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Transcript

Ex 6.3, 8 Find a point on the curve 𝑦=(π‘₯βˆ’2)^2 at which the tangent is parallel to the chord joining the points (2, 0) and (4, 4).Given Curves is 𝑦=(π‘₯βˆ’2)^2 Let AB be the chord joining the Point (2 , 0) & (4 ,4) & CD be the tangent to the Curve 𝑦=(π‘₯βˆ’2)^2 Given that tangent is Parallel to the chord i.e. CD βˆ₯ AB ∴ Slope of CD = Slope of AB If two lines are parallel, then their slopes are equal Slope of tangent CD Slope of tangent CD = 𝑑𝑦/𝑑π‘₯ =(𝑑(π‘₯ βˆ’ 2)^2)/𝑑π‘₯ = 2(π‘₯βˆ’2) (𝑑 (π‘₯ βˆ’ 2))/𝑑π‘₯ = 2(π‘₯βˆ’2) (1βˆ’0) = 2(π‘₯βˆ’2) Slope of AB As AB is chord joining Points (2 , 0) & (4 , 4) Slope of AB =(4 βˆ’ 0)/(4 βˆ’ 2) =4/2 As slope of line joining point (π‘₯ , 𝑦) & (π‘₯2 , 𝑦2) 𝑖𝑠 (𝑦2 βˆ’ 𝑦1)/(π‘₯2 βˆ’ π‘₯1) =2 Now, Slope of CD = Slope of AB 2(π‘₯βˆ’2)=2 π‘₯βˆ’2=2/2 π‘₯βˆ’2=1 π‘₯=3 Finding y when π‘₯=3 𝑦=(π‘₯βˆ’2)^2 𝑦=(3βˆ’2)^2 𝑦=(1)^2 𝑦=1 Hence, Point is (πŸ‘ , 𝟏) Thus, the tangent is parallel to the chord at (3 ,1)

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.